# The proper Schwarzschild radial distance between two spherical shells

• I
Gold Member

## Summary:

I am trying to understand the use of the Schwarzschild metric in measuring the "proper" distance between two non-moving concentric spherical shells which are also both concentric with the event horizon sphere which of course has the Schwarzschild radius.
If what I describe below is correct, I hope someone will confirm this for me. If it is is incorrect, I hope someone with explain my error to me.

## Main Question or Discussion Point

For the purpose of this thread the metric is
ds2 = - (1-rs/r) c2 dt2 + dr2 / (1-rs/r)​
where
rs = 2GM/c2.​
(I modified the above from

I assume that the two spherical shells are stationary. Therefore
dt = 0.​
The r coordinate for the radii of the two shells satisfy the relationships:
A1 = 4 π (r1)2
is the surface area of one spherical shell, and
A2 = 4 π (r2)2
is the area of the other spherical shell.

The proper radial distance D between the r1 shell and the r2 shell is:
D = ∫r1r2 (1/(1-r/rs))1/2 dr.​

My purpose in studying this problem is that I am interested in estimating the error if I use Newtonian gravity equations for orbital motion rather than the more accurate (but more difficult to use) Schwarzschild math.

kent davidge and vanhees71

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Nugatory
Mentor
I am interested in estimating the error if I use Newtonian gravity equations for orbital motion...
Do remember that the trajectory of an orbiting body is not a curve of constant ##t##

Gold Member
Do remember that the trajectory of an orbiting body is not a curve of constant ##t##
Hi Nugatory:

I do understand and that time plays a role in orbital behavior. I plan to limit my study (at least for a while) to radial motion only to keep the math simpler. I am assuming that that the particular approximation of error using Newtonian gravity math rather than Schwartschild math will be similar for time as for distance.

Regards,
Buzz

pervect
Staff Emeritus
The proper radial distance D between the r1 shell and the r2 shell is:
D = ∫r1r2 (1/(1-r/rs))1/2 dr.​

My purpose in studying this problem is that I am interested in estimating the error if I use Newtonian gravity equations for orbital motion rather than the more accurate (but more difficult to use) Schwarzschild math.
I'm not sure I understand your motivations, but I don't see anything wrong with what you actually wrote. It would be cleaner if expressed in Latex, looking like this:

$$D = \int_{r_1}^{r_2} \frac{dr}{\sqrt{1-\frac{r}{r_s}}}$$

and is generated by

Code:
$$D = \int_{r_1}^{r_2} \frac{dr}{\sqrt{1-\frac{r}{r_s}}}$$
https://www.physicsforums.com/help/latexhelp/ has more help on Latex on PF.

Buzz Bloom
PAllen
2019 Award
In Euclidean geometry, given two concentric spheres given areas, there is an implied radial distance between them. In Schwarzschild geometry, the radial distance given two concentric sheres of the same area is different from the Euclidean expectation. What I don't see is what this has to do with calculating orbits, whether you use Newtonian gravity or GR.

PeterDonis
Ibix
I plan to limit my study (at least for a while) to radial motion only to keep the math simpler.
If you are limiting yourself to purely radial motion you can actually solve the differential equations for orbital motion in Schwarzschild coordinates and write ##\tau(r)## and ##t(r)## (unfortunately I don't think you can invert them to get ##r(\tau)## or ##t(\tau)##). These coordinate-based expressions aren't directly useful, but you can derive some direct observables (e.g. time of flight for a thrown ball) and make comparisons between Newtonian and relativistic predictions.

Buzz Bloom
Gold Member
Hi @pervect, @PAllen, and @Ibix:

Regarding the use of Latex, I agree that it is a better form of presentation than what I used, but I have had difficulties using Latex. When I try to preview a post with Latex before posting, I experience display issues that make correcting the Latex very difficult.

Regarding the comparing of coordinate radial distance with proper radial distance, I want to be as certain as I can be that with respect to the distances I want to explore the difference is sufficiently small that it can be reasonably ignored. The problem I want to explore involves calculating whether a test object escapes from a gravitational source or is bound to it, involving a combination of gravitational attraction by a Euclidean point, and the outward acceleration from the expansion of the universe. I feel this topic has not been adequately discussed in several threads, for example.
and

I think that I am now ready to start a new thread specifically to explore (limited to radial motion) of what seems to me (possibly incorrectly) to be a correct way to determining whether a test object moving at a specified radial velocity will escape from a single given mass at a given distance. @PeterDonis has made an good argument that (for practical purposes) the Euclidean escape velocity is a sufficient test, even though it ignores the influence of the expanding universe. I want to explore in detail comparing this practical test with a test including the expanding universe. I have not been able to find on the Internet any source which explores this topic.

Regards,
Buzz

PAllen
2019 Award
First, it is only the acceleration of expansion that will have any effect at all on escape velocity (the first derivative of scale factor has exactly zero tidal effect). Second, for any reasonable universe parameters, this effect will be much smaller than the difference between GR and Newtonian escape velocity. Thus, to analyze this effect, you would need to merge a Schwarzschild solution and an FLRW solution, which is a nontrivial undertaking.

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PeterDonis
PeterDonis
Mentor
2019 Award
@PeterDonis has made an good argument that (for practical purposes) the Euclidean escape velocity is a sufficient test, even though it ignores the influence of the expanding universe.
Where did I make this argument?

Gold Member
Where did I make this argument?
Hi Peter:

I will make a search to find some specific posts. I have forgotten the exact threads where they appear.

Regards,
Buzz

Gold Member
First, it is only the acceleration of expansion that will have any effect at all on escape velocity (the first derivative of scale factor has exactly zero tidal effect). Second, for any reasonable universe parameters, this effect will be much smaller than the difference between GR and Newtonian escape velocity. Thus, to analyze this effect, you would need to merge a Schwarzschild solution and an FLRW solution, which is a nontrivial undertaking.
Hi Paul:

I may well have a misunderstanding regarding your post, and I will reply as soon as I can. I am now off to a meeting.

Regards,
Buzz

Gold Member
Hi @PeterDonis:

I found one post which I quote from below. I still think there were related other posts, but I failed to find any others that were on topic.

Post #57​
... the correct criterion for whether a pair of objects are gravitationally bound is whether their relative recession velocity is greater than escape velocity.​

I understand that the definition you are using for "escape velocity" is the standard Euclidean one:
Ve2 = √(2M/D).​
This ignores the influence of the Hubble acceleration AH:
AH approx = H2D.​
This approximation ignores the changes in H during the time it takes as the test particle moves.I am still working on how to include this change in H, or to determine that the change in H is negligible regarding the issue of boundedness.

Regards,
Buzz

Gold Member
Poking around a bit, it seems that the key thing to search for is the McVittie metric, which is an exact solution for a mass point embedded in an FLRW cosmology. Here are two example references:

https://arxiv.org/abs/1508.04763
https://iopscience.iop.org/article/10.1088/1742-6596/484/1/012044/pdf
Hi Paul:

I looked at the first reference and found some strange discussions. First, the date and title:
(2002) Search for a Standard Explanation of the Pioneer Anomaly.​
I did a search on "Pioneer Anomaly" and found (2012)
Why was the thermal emission from the spacecraft anisotropic and slowing the spacecraft down? First of all, because the Pioneer spacecraft were spin-stabilized and almost always pointed their big dishes towards Earth. Second of all, because two sources of thermal radiation (heat) were then on the leading side of the spacecraft.​

I confess this seems like it is not relevant to the problem I am exploring, but I will give it and the other article a deeper look over the next few days.

Regards,.
Buzz

Ibix
I looked at the first reference and found some strange discussions. First, the date and title:
(2002) Search for a Standard Explanation of the Pioneer Anomaly.
This article would appear to be https://arxiv.org/abs/gr-qc/0107022, while the article @PAllen linked is https://arxiv.org/abs/1508.04763, titled "Lensing in the McVittie metric". Perhaps you confused two different browser tabs?

I have to admit I'm not clear what it is you want to do. Are you just trying to determine the escape velocity from some object in an expanding universe? Then the McVittie metric would appear to be what you want to study, from PAllen's description. In particular you'd want to look at radially outgoing timelike geodesics. I suspect that there are quite a few potentially thorny questions on the way to this, though.

Aside: I could murder a chocolate digestive...

PAllen
PAllen
2019 Award
Hi Paul:

I looked at the first reference and found some strange discussions. First, the date and title:
(2002) Search for a Standard Explanation of the Pioneer Anomaly.​
I did a search on "Pioneer Anomaly" and found (2012)
Why was the thermal emission from the spacecraft anisotropic and slowing the spacecraft down? First of all, because the Pioneer spacecraft were spin-stabilized and almost always pointed their big dishes towards Earth. Second of all, because two sources of thermal radiation (heat) were then on the leading side of the spacecraft.​

I confess this seems like it is not relevant to the problem I am exploring, but I will give it and the other article a deeper look over the next few days.

Regards,.
Buzz
I don’t understand. My link is to: ”Lensing in the McVittie metric”, from 2015. There is a brief mention of the pioneer anomaly, but that is tangential to the paper’s topic. The paper discusses the background of using the McVittie metric to analyze small scale effects of expansion. Its specific focus on lensing is not relevant, but the purpose of referencing it was to answer my own point that you need a way to combine Scwarzschild metric with FLRW metric to answer the questions you pose. After a little research, I found the McVittie metric is exactly such a proposal. This paper presents the McVittie metric with background and many references, thus perfectly on point from this point of view.

PeterDonis
Mentor
2019 Award
I understand that the definition you are using for "escape velocity" is the standard Euclidean one
"Euclidean" is not the right word for the first formula I gave in that post for ##V_E##. That formula is derived from Schwarzschild spacetime; it is the formula for the escape velocity at radial coordinate ##D## in Schwarzschild spacetime relative to a mass ##M##. For large radial coordinates, the radial coordinate is a good approximation to actual radial distance, so ##D## can be considered the distance to the object.

This ignores the influence of the Hubble acceleration
I'm not sure what you mean by "Hubble acceleration", but if you mean the effect of dark energy, go back and read the post you reference again, since the second formula I gave for ##V_E## in that post explicitly takes the effects of dark energy into account. As I noted in the post, this decreases the effective distance ##D## from a given mass at which a comoving object will be moving away from the mass at escape velocity, so if you're just looking for an upper bound to that distance, the first formula is fine.

This approximation ignores the changes in H during the time it takes as the test particle moves
##H## is decreasing with time, so if you want an upper bound, just use the value of ##H## at the end of whatever time period you're interested in, since that will be the lowest value (which will therefore give you the largest value of ##D##).

PeterDonis
Mentor
2019 Award
a way to combine Scwarzschild metric with FLRW metric
Note that the McVittie metric uses isotropic coordinates, so the form of the Schwarzschild metric it is implicitly using is different from the form I used in the post in another thread that @Buzz Bloom linked to and that I was discussing in my previous post. For large distances from a given mass, the difference in radial coordinates between isotropic and Schwarzschild coordinates is negligible, and both are well approximated by the actual proper distance ##D##, so this doesn't significantly affect what I posted, but it's worth being aware of.

Buzz Bloom and PAllen
PeterDonis
Mentor
2019 Award
the McVittie metric uses isotropic coordinates
Btw, another implication of this is that describing the McVittie metric as a "mass point" embedded in an FRW universe is not quite correct. The metric does not cover the spacetime region at or below the event horizon of the mass at all (it can't since isotropic coordinates do not cover that region). So it is better described as a metric for the exterior of a black hole embedded in an FRW universe.

Gold Member
First, it is only the acceleration of expansion that will have any effect at all on escape velocity
Hi Paul:

I do not understand why the above might be true.

Imagine two masses, each of mass M, and each occupying a non-moving point relative to co-moving coordinates. Assume at time t0, the distance between the centers of these two masses is D. The Hubble expansion tells us that at time t0 they are moving apart at velocity HD:
V(t0) = H(t0)D(t0).​
There is also a Hubble acceleration happening. I do not mean the acceleration of the expansion. I mean
A(t) = dV(t)/dt = H dD/dt + D dH(t)/dt.​
If we assume
ΩR = ΩM = Ωk = 0​
and
ΩΛ=1,​
then
dH(t)/dt = 0,​
H(t) = (1/a) d(a(t))/dt,​
and​
a(t) = eHt.​
With these assumptions,
A(t) = HV = H2D.​
I call this the Hubble acceleration, but if that is an ambiguous name, I will have to chose some other name.

Now assume (Newtonian view) the inward gravitational acceleration on a test particle at a distance D from the gravitating point mass M equals the Hubble acceleration.
GM/D2 = H2D​
D = (GM/H2)(1/3).​
The implication is that the test particle with no velocity will be stationary with no acceleration. This is, this particle will be in an "orbit" with zero orbital velocity.

It seems natural to assume that the Hubble acceleration applies to a test particle at a distance D from a massless origin point P, when (1) the test particle is moving away from P at velocity HD, and (2) it is accelerating away from P with accleration H2D. Apparently, there is some doubt that (assuming the Newtonian view) if there is a mass M at the origin P, the test particle would not be influenced by H2D, and it would not have a total acceleration of
A = -GM/D2 + H2D.​

Do you know any reference that explains why the Hubble acceleration affects the motion of a test particle at a distance where the gravitational attraction becomes negligible, but this does not apply when the gravitational attraction has a similar magnitude. I have been unable to find any such explanation anywhere on the Internet.

Regards,
Buzz

PeterDonis
Mentor
2019 Award
I do not understand why the above might be true.
It's because expansion (as distinct from accelerated expansion) is just inertia; faraway comoving objects are moving away from us because of inertia. That fact cannot affect the escape velocity for objects from, say, the Milky Way.

A slightly more technical version of the above is that, because of the Shell Theorem, a spherically symmetric distribution of matter surrounding some region cannot affect the motion of anything inside the region. To a very good approximation, the distribution of matter in the rest of the universe with respect to a single gravitationally bound system like the Milky Way is spherically symmetric, so it can be ignored when analyzing the motion of objects within the system. The same applies to analyzing what the escape velocity from the system would be.

Also, please learn how to use the PF LaTeX feature. Most of the rest of your post is basically unreadable because of the equation soup and the lack of proper formatting.

PeterDonis
Mentor
2019 Award
There is also a Hubble acceleration happening. I do not mean the acceleration of the expansion.
Yes, you do. In the formula you derive, you explicitly ignore everything except ##\Omega_{\Lambda}##, which is dark energy, i.e., "acceleration of the expansion".

I already explicitly acknowledged, in the previous thread you linked to, that the presence of dark energy does affect escape velocity, and gave you the correct formula for escape velocity in the presence of dark energy and how that formula is derived.

PAllen
2019 Award
@Buzz Bloom , the proof of my statement is simple. Two different arguments are possible. First, only curvature, as an invariant object, can affect physics, and this depends on second derivatives of the metric components, which means only second derivatives of scale factor can contribute to tidal gravity. A more cute argument is to ask what happens if the derivative of the scale factor is constant. The answer is that the curvature vanishes identically, which means you have flat Minkowski space in funny coordinates (Milne coordinates). Choice of funny coordinates cannot possibly affect orbits or escape velocity, this linear expansion rate must have exactly zero local effect.

In fact, it is well known that the following possibly counterintuitive statements are true, that follow from the fact that only the second derivative of scale factor matter for local physics:

- if you have a forever expanding universe, such that the rate of expansion is forever decreasing, then local orbits will be smaller than expected, and escape velocity larger.

-
if you have a forever contracting universe, where the rate of contraction is forever decreasing, then local orbits will be larger than expected, and escape velocity smaller than expected.

Of course, our universe is presumed to have accelerated expansion, so escape velocity will be smaller than expected, and orbits larger than expected.

Buzz Bloom
Gold Member
lack of proper formatting
Hi Peter:

I will work on getting familiar with Latex, but I fear that my frustration limit is likely to prevent me from using it regularly.

Aside from no Latex, can you give me some examples of what you mean by my "lack of proper formatting"?

Regards,
Buzz

Gold Member
In fact, it is well known that the following possibly counterintuitive statements are true, that follow from the fact that only the second derivative of scale factor matter for local physics:
Hi Paul:

I think it would be very helpful to me to see a lengthy explanation of
"only the second derivative of scale factor matter for local physics". Can you recommend a paper or book that explains the physical and/or mathematical reason for this limitation.

Regards,
Buzz

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